Question

After being poured into a pan, 10.0 moles of water were heated on an electric stove. During this time, the temperature of the liquid rose from 26 ℃ to 38 ℃. How much heat, in J, was absorbed by the water? Use c = 73.32 J/mol K.

Answer 1

$Q=cn\Delta{T}=73.32\frac{J}{mol×K}×10.0\ mol×(38°C-26°C)=8800\ J$

Answer: 8800 J